# Thread: Work problem- draining a pool

1. ## Work problem- draining a pool

I did a similar problem and got it right but I can't seem to get this right:

You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved perfectly into the water. However, now that you want to swim, you must remove all of the Kool-Aid contaminated water. The swimming pool is round, with a 10 foot radius. It is 11.5 feet tall and has 3 feet of water in it.
How much work is required to remove all of the water by pumping it over the side? Use the physical definition of work, and the fact that the weight of the Kool-Aid contaminated water is .

So I know that Volume= 100 pi dy

Mass=100pi* dy * 65lb/ft^3

Force = 100pi* dy * (65lb/ft^3) * 9.8 (gravity)
So Work= 100pi* dy * (65lb/ft^3) * 9.8 (gravity) (11.5-y)

work=63700 pi intregal (11.5-y)dy ( from 0 to 3)

work=63700pi (11.5y-(y^2)/2) ( from 0 to 3)

63700pi *30 lb/ft

What did I do wrong?

2. Because this is a cylinder, everything is "linear" and you don't need Calculus. It is easy to find the center of gravity of the water. Taking a coordinate system with origin at the center of the bottom of the pool, and the positive z-axis upward, and units of feet, the center of gravity of the water is at (0, 0, 1.5). That means that the water, volume $\pi (5)^2(3)= 75\pi\approx 235.6$ cubic feet and so (235.6)(65)= 15315 pounds must be lifted an average of 11,5- 1.5= 10 feet. The work necessary is 15315(10)= 153150 foot-pounds.

Your error, I believe, was multiplying by 9.8. That is the acceleration due to gravity in the metric system and multiplying by it will convert mass to weight in the metric system. But here, you were given the weight density of water, not mass density, as pounds per cubic foot, in the English system.

3. Thanks! Yeah, now that I looked at the other problem I got right, I saw that the previous question I got right was in metrics.

Now, is it possible to do this problem using integrals? Since this is a calculus class, I was thinking there is a way to do it that way right? Only reason I am asking is in case the exam say "use integrals"

p.s. one tiny mistype you made was that the radius is actually 10, you had it as 5. But I got the gist! Thanks!

4. Yes, of course, my point was that you don't need Calculus. (Just as you could use Calculus to find the rate of change of y= 3x+ 2, but you don't need to.)

A "layer of water" at height x feet above the ground has to be lifted a distance 11.5- x feet. It is a disk of area $25\pi$ square feet and, with thickness, dx, volume $25\pi dx$ cubic feet and so weighing $(65)(25)\pi dx= 1625\pi dx$ pounds. The work done lifting that out of the pool is $1625\pi (11.5- x)dx$ foot-pounds.

The integral you want is $1625\pi \int_0^3(11.5- x)dx$
(Amazingly, when I do the0 problem that way, I get the same answer as I did before.)

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### swimming pool work problem ca

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